\(=\left(\dfrac{2sinx.cosx}{cos2x}-\dfrac{sinx}{cosx}\right)\left(2sinx.cosx-\dfrac{sinx}{cosx}\right)\)

\(=sinx\left(\dfrac{2cosx}{cos2x}-\dfrac{1}{cosx}\right).sinx\left(2cosx-\dfrac{1}{cosx}\right)\)

\(=sin^2x\left(\dfrac{2cos^2x-\left(2cos^2x-1\right)}{cosx.cos2x}\right)\left(\dfrac{2cos^2x-1}{cosx}\right)\)

\(=sin^2x\left(\dfrac{1}{cosx.cos2x}\right)\left(\dfrac{cos2x}{cosx}\right)=\dfrac{sin^2x}{cos^2x}=tan^2x\)

\(cotx-tanx=\frac{cosx}{sinx}-\frac{sinx}{cosx}=\frac{cos^2x-sin^2x}{sinx.cosx}=\frac{cos2x}{\frac{1}{2}sin2x}=2cot2x\)

\(\frac{cos^2x-sin^2x}{1+sin2x}=\frac{\left(cosx-sinx\right)\left(cosx+sinx\right)}{sin^2x+cos^2x+2sinx.cosx}=\frac{\left(cosx-sinx\right)\left(cosx+sinx\right)}{\left(cosx+sinx\right)^2}=\frac{cosx-sinx}{cosx+sinx}\)

\(=\frac{\frac{cosx}{cosx}-\frac{sinx}{cosx}}{\frac{cosx}{cosx}+\frac{sinx}{cosx}}=\frac{1-tanx}{1+tanx}\)

\(\frac{1-sinx-cos2x}{sin2x-cosx}=\frac{1-sinx-\left(1-2sin^2x\right)}{2sinxcosx-cosx}=\frac{2sin^2x-sinx}{2sinxcosx-cosx}\)

\(=\frac{sinx\left(2sinx-1\right)}{cosx\left(2sinx-1\right)}=\frac{sinx}{cosx}=tanx\)

a: \(VT=\dfrac{cot^2x}{1+cot^2x}\cdot\dfrac{1+tan^2x}{tan^2x}\)

\(=\dfrac{cot^2x}{\dfrac{1}{sin^2x}}\cdot\dfrac{\dfrac{1}{cos^2x}}{tan^2x}\)

\(=\dfrac{cot^2x}{tan^2x}\cdot\dfrac{1}{cos^2x}:\dfrac{1}{sin^2x}\)

\(=\dfrac{cot^2x}{tan^2x}\cdot\dfrac{sin^2x}{cos^2x}\)

\(=cot^2x\)

\(VP=\dfrac{tan^2x+cot^2x}{1+tan^4x}=\dfrac{\dfrac{sin^2x}{cos^2x}+\dfrac{cos^2x}{sin^2x}}{1+\dfrac{sin^4x}{cos^4x}}\)

\(=\dfrac{sin^4x+cos^4x}{sin^2x\cdot cos^2x}:\dfrac{cos^4x+sin^4x}{cos^4x}\)

\(=\dfrac{sin^4x+cos^4x}{sin^2x\cdot cos^2x}\cdot\dfrac{cos^4x}{cos^4x+sin^4x}=\dfrac{cos^2x}{sin^2x}=cot^2x\)

=>VT=VP

b:

\(\dfrac{tan^2x-cos^2x}{sin^2x}+\dfrac{cot^2x-sin^2x}{cos^2x}\)

\(=\dfrac{\left(\dfrac{sinx}{cosx}\right)^2-cos^2x}{sin^2x}+\dfrac{\left(\dfrac{cosx}{sinx}\right)^2-sin^2x}{cos^2x}\)

\(=\dfrac{sin^2x-cos^4x}{cos^2x\cdot sin^2x}+\dfrac{cos^2x-sin^4x}{sin^2x\cdot cos^2x}\)

\(=\dfrac{sin^2x+cos^2x-cos^4x-sin^4x}{cos^2x\cdot sin^2x}\)

\(=\dfrac{1-\left(cos^2x+sin^2x\right)^2+2\cdot cos^2x\cdot sin^2x}{cos^2x\cdot sin^2x}\)

\(=\dfrac{2\cdot cos^2x\cdot sin^2x}{cos^2x\cdot sin^2x}=2\)

\(2sin\left(\frac{\pi}{4}+a\right)sin\left(\frac{\pi}{4}-a\right)=cos2a-cos\left(\frac{\pi}{2}\right)=cos2a\)

\(tanx-\frac{1}{tanx}=\frac{sinx}{cosx}-\frac{cosx}{sinx}=\frac{sin^2x-cos^2x}{sinx.cosx}=-\frac{2\left(cos^2x-sin^2x\right)}{2sinx.cosx}=\frac{2cos2x}{sin2x}=-2cot2x=-\frac{2}{tan2x}\)

\(sinx\left(1+cos2x\right)=sinx\left(1+2cos^2x-1\right)=2sinx.cosx.cosx=sin2x.cosx\)

\(tanx-\frac{1}{tanx}=\frac{sinx}{cosx}-\frac{cosx}{sinx}=\frac{sin^2x-cos^2x}{sinx.cosx}=\frac{-cos2x}{\frac{1}{2}sin2x}=-\frac{2}{tan2x}\)

\(tan\frac{x}{2}\left(\frac{1}{cosx}+1\right)=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}\left(\frac{1+cosx}{cosx}\right)=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}.\frac{2cos^2\frac{x}{2}}{cosx}=\frac{2sin\frac{x}{2}.cos\frac{x}{2}}{cosx}=\frac{sinx}{cosx}=tanx\)

\(\frac{sin2x-sin4x}{1-cos2x+cos4x}=\frac{sin2x-2sin2x.cos2x}{1-cos2x+2cos^22x-1}=\frac{sin2x\left(1-2cos2x\right)}{-cos2x\left(1-2cos2x\right)}=\frac{-sin2x}{cos2x}=-tan2x\)

\(\frac{sin4x-sin2x}{1-cos2x+cos4x}=-\left(\frac{sin2x-sin4x}{1-cos2x+cos4x}\right)=-\left(-tan2x\right)=tan2x\) lấy luôn kết quả câu trên cho lẹ, biến đổi thì làm y hệt

\(\left(\frac{1}{cos2x}+1\right)tanx=\left(\frac{cos2x+1}{cos2x}\right).\frac{sinx}{cosx}=\frac{2cos^2x}{cos2x}.\frac{sinx}{cosx}\)

\(=\frac{2sinx.cosx}{cos2x}=\frac{sin2x}{cos2x}=tan2x\)

\(\frac{cos7a+cosa+cos5a+cos3a}{sin7a+sina+sin5a+sin3a}=\frac{2cos4a.cos3a+2cos4a.cosa}{2sin4a.cos3a+2sin4a.cosa}\)

\(=\frac{cos4a\left(2cos3a+2cosa\right)}{sin4a\left(2cos3a+2cosa\right)}=\frac{cos4a}{sin4a}=cot4a\)